2023 AIME I Problems/Problem 13
Problem
Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths and . The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is , where and are relatively prime positive integers. Find . A parallelepiped is a solid with six parallelogram faces such as the one shown below.
{insert diagram here}
Solution 1 (3-D Vector Analysis)
Denote . Denote by the length of each side of a rhombus.
Now, we put the solid to the 3-d coordinate space. We put the bottom face on the plane. For this bottom face, we put a vertex with an acute angle at the origin, denoted as . For two edges that are on the bottom face and meet at , we put one edge on the positive side of the -axis. The endpoint is denoted as . Hence, . We put the other edge in the first quadrant of the plane. The endpoint is denoted as . Hence, .
For the third edge that has one endpoint , we denote by its second endpoint. We denote . Without loss of generality, we set . Hence,
We have and
Case 1: or .
By solving (2) and (3), we get
Plugging these into (1), we get
Case 2: and , or and .
By solving (2) and (3), we get
Plugging these into (1), we get
We notice that . Thus, (4) (resp. (5)) is the parallelepiped with a larger (resp. smaller) height.
Therefore, the ratio of the volume of the larger parallelepiped to the smaller one is
Recall that . Thus, . Plugging this into the equation above, we get
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (no trig)
Let one of the vertices be at the origin and the three adjacent vertices be , , and . For one of the parallelepipeds, the three diagonals involving the origin have length . Hence, and . Since all of , , and have equal length, , , and . Symmetrically, , , and . Hence the volume of the parallelepiped is given by .
For the other parallelpiped, the three diagonals involving the origin are of length and the volume is .
Consequently, the answer is , giving .
~EVIN-
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